# Grignard reagent reaction mechanism: Which bond breaks during the decomposition of the adduct?

We just finished our Organic Chem. chapter Aldehydes, Ketones and Carboxylic acids at school, but I still have one lingering doubt.

Considering the reaction between an alkyl magnesium halide and an aldehyde or ketone: (Diagram's from Wikipedia)

Now my teacher tells me that in the adduct, it's the Mg-O bond that gets cleaved during hydrolysis and not the C-O bond, resulting in an alcohol.

He isn't willing to elaborate ("Aaron, it happens... just like that... you shouldn't worry about all the complexities behind everything...") and I can't seem to find any source on the web that deals with the particulars of bond-breakage.

I have doubts regarding that point of his, since the Mg-O bond has significant ionic character (due to the electronegativity difference between the atoms. But in comparison, the C-O bond is more covalent, and therefore must be easier to hydrolyze. Because (the way my high-schooler brain sees it) ionic bonds are established by strong electrostatic forces of attraction between the ions concerned, while this isn't the case for predominantly covalent bonds. So it seems to make sense that the weaker bond is preferably hydrolyzed, hence I'm mooting for the C-O hydrolysis.

Who's right here? Could someone cite any (reliable) sources?

• Why should a covalent bond be easier to hydrolyse? If you insist on making any generalisation at all, surely it should be the other way round. Just look at NaI (ionic) vs AgI (covalent-ish). Water can only break one of those bonds. Oct 8 '16 at 8:38

As can be seen above, the end product is the same whether we split the $\ce{Mg-O}$ bond or the $\ce{C-O}$ bond.
• (If I went out of my way, I would even propose a cyclic rearrangment where the $\ce{H-OH}$ is added to the $\ce{O-Mg}$ bond.)